# How do you graph r=2+sin theta?

Oct 28, 2016

In $\left[0. \pi\right]$, the graph crowns over the circle r = 2. In $\left[\pi , 2 \pi\right]$, it hangs from the circle. The maximum distance from the circle, either way, is 1 unit.

#### Explanation:

$r \left(\theta\right)$ is periodic, with period $2 \pi$ in $\theta$.

The Table for one period theta in [0, 2pi] is sufficient.

$\left(r , \theta\right)$:

$\left(2 , 0\right) , \left(2 + \frac{1}{\sqrt{2}} , \frac{\pi}{4}\right) \left(3 , \frac{\pi}{2}\right) , \left(2 + \frac{1}{\sqrt{2}} , \frac{3}{4} \pi\right) \left(2 , \pi\right)$

$\left(2 - \frac{1}{\sqrt{2}} , \frac{5}{4} \pi\right) \left(1 , \frac{3}{2} \pi\right) \left(2 - \frac{1}{\sqrt{2}} , \frac{7}{4} \pi\right) \left(2 , 2 \pi\right)$

Altogether this graph is a wave, twining around a circle.

graph{x^2 + y^2 - 2sqrt(x^2 + y^2) - y = 0}

Credit for the graphs goes to Socratic.

graph{(x^2 + y^2)^1.5 - 2(x^2 + y^2) - 2xy = 0}

The second graph is for $r = 2 + \sin \left(2 \theta\right)$

graph{(x^2 + y^2)^2 - 2(x^2 + y^2)^1.5 - 3(x^2y-xy^2) = 0}

The third is for $r = 2 + \sin \left(3 \theta\right)$

graph{(x^2 + y^2)^2.5 - 2(x^2 + y^2)^2 -4(x^3y-xy^3) = 0}

The fourth is for $r = 2 + \sin \left(4 \theta\right)$